MAT01A1: Hyperbolic Functions

Dr Craig

Week: 11 May 2020

This is a short “lecture” on an important

type of functions: the hyperbolic functions.

Hyperbolic functions have many applications

in physics, engineering, and applied

mathematics.

Please note that we only cover pages

159–161 of this section. We do not cover

inverse hyperbolic functions.

You can of course look over the inverse

functions and their derivatives on your own.

Hyperbolic functions

The hyperbolic functions are made by

combining ex and e−x in different ways.

They are related to hyperbolas similarly to

how trig functions are related to circles.

This is why their names are similar to trig

functions.

However, you need to be careful: many of

the identities and derivatives look like the

corresponding ones for trig functions, but

they are slightly different.

The main two hyperbolic functions:

sinh(x) =ex − e−x

2cosh(x) =

ex + e−x

2

The full name of the first function is

“hyperbolic sine” while the second one is

“hyperbolic cosine”.

sinhx is often read as “shine x”

coshx is often read as “cosh x”

The main two hyperbolic functions:

sinh(x) =ex − e−x

2cosh(x) =

ex + e−x

2

The full name of the first function is

“hyperbolic sine” while the second one is

“hyperbolic cosine”.

sinhx is often read as “shine x”

coshx is often read as “cosh x”

Definition of the Hyperbolic Functions:

sinh(x) =ex − e−x

2csch(x) =

1

sinhx

cosh(x) =ex + e−x

2sech(x) =

1

coshx

tanh(x) =sinhx

coshxcoth(x) =

coshx

sinhx

ran(sinh(x)) = (−∞,∞)

ran(cosh(x)) = [1,∞)

Remembering the definitions

y = c+ a cosh(x/a)

The cosh function is used to describe, for instance,a chain hanging between two fixed points. Youshould know the shape of the two graphs and thefunction definitions. If you think coshx↔ chainthen you will remember which one is which.

Hyperbolic Identities:

sinh(−x) = − sinh(x) cosh(−x) = cosh(x)

cosh2(x)− sinh2(x) = 1

1− tanh2(x) = sech2(x)

sinh(x + y) = sinhx cosh y + coshx sinh y

cosh(x + y) = coshx cosh y + sinhx sinh y

The proofs of all the identities follow directly

from the definition of cosh(x) and sinh(x).

Hyperbolic Identities:

sinh(−x) = − sinh(x) cosh(−x) = cosh(x)

cosh2(x)− sinh2(x) = 1

1− tanh2(x) = sech2(x)

sinh(x + y) = sinhx cosh y + coshx sinh y

cosh(x + y) = coshx cosh y + sinhx sinh y

The proofs of all the identities follow directly

from the definition of cosh(x) and sinh(x).

Proofs of identities:

On the next slide we will prove

cosh(x + y) = coshx cosh y + sinhx sinh y

Use the proof there as a guide to writing up

your own proof for

sinh(x + y) = sinhx cosh y + coshx sinh y

Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)

2

We have

RHS =

(ex + e−x

2

)(ey + e−y

2

)+

(ex − e−x

2

)(ey − e−y

2

)

=ex+y + ex−y + ey−x + e−x−y

4+

ex+y − ex−y − ey−x + e−x−y

4

=2ex+y + 2e−x−y

4=

ex+y + e−(x+y)

2= LHS

Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)

2We have

RHS =

(ex + e−x

2

)(ey + e−y

2

)+

(ex − e−x

2

)(ey − e−y

2

)

=ex+y + ex−y + ey−x + e−x−y

4+

ex+y − ex−y − ey−x + e−x−y

4

=2ex+y + 2e−x−y

4=

ex+y + e−(x+y)

2= LHS

Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)

2We have

RHS =

(ex + e−x

2

)(ey + e−y

2

)+

(ex − e−x

2

)(ey − e−y

2

)

=ex+y + ex−y + ey−x + e−x−y

4+

ex+y − ex−y − ey−x + e−x−y

4

=2ex+y + 2e−x−y

4=

ex+y + e−(x+y)

2= LHS

Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)

2We have

RHS =

(ex + e−x

2

)(ey + e−y

2

)+

(ex − e−x

2

)(ey − e−y

2

)

=ex+y + ex−y + ey−x + e−x−y

4+

ex+y − ex−y − ey−x + e−x−y

4

=2ex+y + 2e−x−y

4

=ex+y + e−(x+y)

2= LHS

Substituting gives LHS = cosh(x+y) =ex+y + e−(x+y)

2We have

RHS =

(ex + e−x

2

)(ey + e−y

2

)+

(ex − e−x

2

)(ey − e−y

2

)

=ex+y + ex−y + ey−x + e−x−y

4+

ex+y − ex−y − ey−x + e−x−y

4

=2ex+y + 2e−x−y

4=

ex+y + e−(x+y)

2= LHS

Derivatives of hyperbolic functions

In order to find the derivatives of both sinhx

and coshx, you need the following important

example of the chain rule at work.

d

dx

(e−x

)= e−x · d

dx(−x) = −e−x

Now, as an exercise, use this to calculate the

derivative of both sinhx and coshx. The

final answers for both are on the next slide.

Derivatives of hyperbolic functions

In order to find the derivatives of both sinhx

and coshx, you need the following important

example of the chain rule at work.

d

dx

(e−x

)= e−x · d

dx(−x) = −e−x

Now, as an exercise, use this to calculate the

derivative of both sinhx and coshx. The

final answers for both are on the next slide.

Derivatives of hyperbolic functions

In order to find the derivatives of both sinhx

and coshx, you need the following important

example of the chain rule at work.

d

dx

(e−x

)= e−x · d

dx(−x) = −e−x

Now, as an exercise, use this to calculate the

derivative of both sinhx and coshx. The

final answers for both are on the next slide.

The derivative of both sinh(x) and cosh(x) can becalculated using just their definition and the chainrule. The rest of the derivatives can be calculatedusing the quotient rule and/or chain rule.

The derivative of both sinh(x) and cosh(x) can becalculated using just their definition and the chainrule. The rest of the derivatives can be calculatedusing the quotient rule and/or chain rule.

Combining the derivatives from the previous

slide with our existing differentiation

techniques we can attempt the more

complicated examples on the next slide.

Examples:

1. Find y′ if y = ln(coshx).

2. Find g′(x) if g(x) = x sinhx− coshx.

3. Find f ′(t) if f (t) = arctan(sinh t).

Attempt these on your own. The solutions

are on the slides that follow.

You should aim to simplify them as far as

possible.

Solutions:

(1.) We apply the chain rule to get:

y′ =1

coshx· ddx

(cosh) =sinhx

coshx= tanhx

(2.) We use the product rule:

g′(x) = (d

dx(x) sinhx + x

d

dx(sinhx))− d

dx(coshx)

= sinhx + x coshx− sinhx

= x coshx

Solutions:

(1.) We apply the chain rule to get:

y′ =1

coshx· ddx

(cosh) =sinhx

coshx= tanhx

(2.) We use the product rule:

g′(x) = (d

dx(x) sinhx + x

d

dx(sinhx))− d

dx(coshx)

= sinhx + x coshx− sinhx

= x coshx

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2.

Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t) =

cosh t

1 + sinh2 t

=cosh t

cosh2 t=

1

cosh t

= sech t

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2. Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t)

=cosh t

1 + sinh2 t

=cosh t

cosh2 t=

1

cosh t

= sech t

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2. Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t) =

cosh t

1 + sinh2 t

=cosh t

cosh2 t=

1

cosh t

= sech t

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2. Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t) =

cosh t

1 + sinh2 t

=cosh t

cosh2 t

=1

cosh t

= sech t

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2. Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t) =

cosh t

1 + sinh2 t

=cosh t

cosh2 t=

1

cosh t

= sech t

Solutions continued...

(3.) We have f (t) = arctan(sinh t). Recall

thatd

dx(arctanx) =

1

1 + x2. Therefore

f ′(t) =1

1 + (sinh t)2· ddt(sinh t) =

cosh t

1 + sinh2 t

=cosh t

cosh2 t=

1

cosh t

= sech t

Prescribed tut problems:

Complete the following exercises from the

8th edition (Ch 3.11):

I 2, 10, 11, 17, 23, 34, 40, 41, 47, 52

I If you are using the 7th edition then do

the numbers above except for the

following changes:

34 → 33, 40 → 38, 41 → 39

Do these exercises before you start on the

slides for Ch 4.4.